How To Draw 450 Degree Angle
Ex xi.1, two - Affiliate xi Form ix Constructions (Term 2)
Terminal updated at March 13, 2021 by Teachoo
Transcript
Ex xi.1, 2 Construct an angle of 45° at the initial indicate of a given ray and justify the structure . Steps of construction Describe a ray OA. Taking O equally eye and any radius, draw an arc cut OA at B. 3. At present, taking B as eye and with the aforementioned radius equally before, draw an arc intersecting the previously fatigued arc at point C. 4. With C as middle and the aforementioned radius, draw an arc cutting the arc at D 5. With C and D as centers and radius more than than 1/ii CD, draw two arcs intersecting at P. 6. Join OP. Thus, ∠ AOP = ninety° Now we depict bisector of ∠ AOP vii. Allow OP intersect the original arc at signal Q eight. Now, taking B and Q every bit centers, and radius greater than 1/2 BQ, describe 2 arcs intersecting at R. nine. Join OR. Thus, ∠ AOR = 45° Justification Nosotros need to prove ∠ AOR = 45° Bring together OC & OB Thus, OB = BC = OC ∴ Δ OCB is an equilateral triangle ∴ ∠ BOC = 60° Bring together OD, OC and CD Thus, OD = OC = DC ∴ Δ Doc is an equilateral triangle ∴ ∠ DOC = 60° Join PD and PC Now, In Δ ODP and Δ OCP OD = OC DP = CP OP = OP ∴ Δ ODP ≅ Δ OCP ∴ ∠ DOP = ∠ COP And then, nosotros can say that ∠ DOP = ∠ COP = 1/two ∠ Medico ∠ DOP = ∠ COP = 1/ii × lx° = 30° (Radius of same arcs) (Arc of aforementioned radii) (Common) (SSS Congruency) (CPCT) (We proved earlier that ∠ Doc= threescore° ) Now, ∠ AOP = ∠ BOC + ∠ COP ∠ AOP = 60° + 30° = 90° At present, Join QR and BR In Δ OQR and Δ OBR OQ = OB QR = BR OR = OR ∴ Δ OQR ≅ Δ OBR ∴ ∠ QOR = ∠ BOR (Radius of same arcs) (Arc of same radii) (Common) (SSS Congruency) ∠ QOR = ∠ BOR = 1/2 ∠ AOP ∠ DOP = ∠ COP = 1/2 × xc° = 45° Thus, ∠ AOR = 45° Hence justified
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